\(\left(x+\dfrac{1}{2}\right).\left(x-\dfrac{3}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
tìm x biết
\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(\left(0.6-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(\left(3x-1\right).\left(-\dfrac{1}{2}x+5\right)=0\)
\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
Tìm x biết :
1/\(\dfrac{4}{3}\) + x = \(\dfrac{2}{15}\) 2/\(\dfrac{1}{3}+\dfrac{1}{2}:\) x = -0,25 3/(\(\dfrac{2x}{5}+2\) ):(-4) = -1\(\dfrac{1}{2}\) 4/ x.( x + 1) = 0 5/3\(\dfrac{1}{2}-\dfrac{1}{2}x\) = \(\dfrac{2}{3}\)
6/1\(\dfrac{1}{2}+x=\dfrac{3}{2}-7\) 7/\(2x-16=40+x\) 8/\(x:\dfrac{5}{3}+\dfrac{1}{3}=\dfrac{-2}{5}\) 9/\(\dfrac{x}{4}=\dfrac{9}{x}\)
\(\left(2x-3\right)\left(6-2x\right)=0\)
\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\dfrac{x}{5}=\dfrac{2}{3}\)
\(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\dfrac{x-12}{4}=\dfrac{1}{2}\)
tìm x biết
a) \(\dfrac{2x-3}{3}+-\dfrac{3}{2}=\dfrac{5-2x}{6}-\dfrac{1}{2}\)
b) \(\left(3x-\dfrac{1}{2}\right).\left(3-\dfrac{1}{2}x\right).\left(1+\dfrac{3}{4}:x\right)=0\)
c) \(\left(3-\dfrac{9}{10}-|x+2|\right):\left(\dfrac{19}{20}-1-\dfrac{2}{3}\right)+\dfrac{4}{5}=1\)
\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(60\%x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\left(2x+\dfrac{3}{2}\right)^2-\dfrac{9}{25}=0\)
\(60\%x+0,4x+x:=2\)
\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)
\(-2x-\dfrac{-3}{5}:\left(-0,5\right)^2=-1\dfrac{1}{4}\)
tìm x:
a)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
b) \(\left(x.6\dfrac{2}{7}+\dfrac{3}{7}\right).2\dfrac{1}{5}-\dfrac{3}{7}=-2\)
c) \(x.3\dfrac{1}{4}+\left(\dfrac{-7}{6}\right).x-1\dfrac{2}{3}=\dfrac{5}{12}\)
d) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
Bài 1:
a) \(\left[\dfrac{3}{20}-\dfrac{1}{5}x\right].1\dfrac{2}{3}=1\dfrac{1}{4}\)
b)\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
c)\(\dfrac{-2}{3} \)-\(\dfrac{1}{3}\left(2x-7\right)=\dfrac{3}{2}\)
\(a,-\dfrac{2}{3}.\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}.\left(2x-1\right)\)
\(b,\dfrac{1}{5}.2^x+\dfrac{1}{3}.2^{x+11}=\dfrac{1}{5}.2^7+\dfrac{1}{3}.2^8\)
\(c,\dfrac{x-19}{345}+\dfrac{x-20}{346}=\dfrac{x-1}{347}+\dfrac{x-22}{348}\)
a) \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{-1}{2}\)
b) \(\left(\dfrac{1}{2}-x\right).\dfrac{2}{3}=\dfrac{1}{8}\)
c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b,\(60\%.x+\dfrac{2}{3}.x=\dfrac{1}{3}.6\dfrac{1}{3}\)
c, \(\left(\dfrac{9}{2}.x-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
d,\(\left(x+\dfrac{5}{6}\right).2\dfrac{2}{5}-1\dfrac{1}{4}=35\%\)
